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• 11 Cuchulainn
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### Re: About solving a transport equation

I’m referring to the pde.

$z=\frac{1}{y-1}$
Why? What's the rationale?

$\frac{\partial u}{\partial t} - (1-y)^2\frac{\partial u}{\partial y} = 0$

Yes, I tried that and get $\frac{\partial u}{\partial t} - \frac{\partial u}{\partial z} = 0$ to get simpler characteristics?

So, do we get $u(z,t) = f(z+t)$?

But it did not seem to be going anywhere. Hold on a sec, I want to check the pde(s) again!! Paul
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### Re: About solving a transport equation

Might be time to hang up your sliderule and leave this to us younger guys!

But, yes, correct.

There might be some tidying up to do, but now the singularity is at least at infinity! Alan
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### Re: About solving a transport equation

The singularity is not a problem. Daniel's Jan 23 post specified the spatial domain as $y \in (0,1)$. Stick to that and there are no issues (see my posted solution). Paul
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### Re: About solving a transport equation

I was trying to point out that a mountain was being made out of a molehill!

The original problem is messier. And to make it slightly interesting have y between plus and minus infinity. Alan
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### Re: About solving a transport equation

Well, I'll agree that it's a molehill (for the OP, who is long gone, I thought it was homework).
But, the (new) original problem is straightforward to solve and not messy at all IMO.
Last edited by Alan on January 28th, 2020, 10:29 pm, edited 2 times in total. Cuchulainn
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### Re: About solving a transport equation

The singularity is not a problem. Daniel's Jan 23 post specified the spatial domain as $y \in (0,1)$. Stick to that and there are no issues (see my posted solution).
Indeed.
The $z$ transformation doesn't work. In fact it corresponds to the pde $\frac{\partial u}{\partial t} - \frac{\partial u}{\partial z} = 0$ on a quarter plane. Then use the transformation $y = z/(z+1)$

It is like going round in circles.
It's not for nothing that $0 < y < 1$ and the pde is a simplified model.
Last edited by Cuchulainn on January 28th, 2020, 10:41 pm, edited 1 time in total. Cuchulainn
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### Re: About solving a transport equation

The original problem is messier. And to make it slightly interesting have y between plus and minus infinity.
Yes, but those are the requirements, They cannot be changed.
What I'm mostly interested in is numerics and most methods flop. Have a look at the Cheyette thread and see for yourself. That's the point.

“The heart of mathematics consists of concrete examples and concrete problems. Big general theories are usually afterthoughts based on small but profound insights; the insights themselves come from concrete special cases.”
Paul Halmos Paul
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### Re: About solving a transport equation

Mostly rubbish!

Exercises for you:

A) Plot the characteristics in z and t
B) Plot in y and t
C) Discuss what happens if you have an initial condition beyond y=1
D) Do similar for the original problem (obvs a different z)
E) Change the direction of time. (You can go back to the (y-1)^2 case for this)
F) How does the solution behave across y=1?
G) Change the y function so that the characteristics hit y=1 in a finite time
H) Now think about the numerics!

Apologies if I am trying to make things interesting!!! Paul
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### Re: About solving a transport equation

But, the (new) original problem is straightforward to solve and not messy at all IMO.
No, it's not at all. If I saw this on an exam it would be the first question I'd go to! Cuchulainn
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### Re: About solving a transport equation

Mostly rubbish!

Exercises for you:

A) Plot the characteristics in z and t
B) Plot in y and t
C) Discuss what happens if you have an initial condition beyond y=1
D) Do similar for the original problem (obvs a different z)
E) Change the direction of time. (You can go back to the (y-1)^2 case for this)
F) How does the solution behave across y=1?
G) Change the y function so that the characteristics hit y=1 in a finite time
H) Now think about the numerics!

Apologies if I am trying to make things interesting!!!
Could do. Quite easy. It is outside the scope at the moment.

H) is the one I am interested in, Alan has provided a solution and I have tested various schemes. I would switch A) and H)  in order of execution.

Using z leads to a singularity, so there is no point. It doesn't work (for my requirements). Can you provide me with a solution $u(z,t)$ in which case questions A) - g) can easily be answered and I can test my schemes against it.

What I have not made explicit is that a domain transformation has been performed, y = 1 is infinity, In fact, this what @macrovue was also doing without telling us. And it didn't help that he deleted the contents of his posts..

I) Generalise (Alan's remark) to
$u_t = b(y) u_y - c(y) u + k(y,t)$ with $u(y,0) = f(y)$.

J) Embed all the savvy into a 1 1/2 factor solver. (don't say it's 1st year undergrad). Paul
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### Re: About solving a transport equation

Alan’s solution is correct. My complaint is that you are both making this far harder than it need be. That’s because the coefficient is only a function of y. So you don’t need the full characteristic analysis (other than to understand what is going on).

The z transformation gives you the same answer!!!

(Perhaps first term second year!) Cuchulainn
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### Re: About solving a transport equation

The z transformation gives you the same answer!!!

That's the question and not all the things we are doing wrong. I usually put it in the Thread of Silly Questions. Paul
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### Re: About solving a transport equation

$u=f\left(1+\frac{y-1}{1-t(y-1)}\right)$ Cuchulainn
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### Re: About solving a transport equation

$u=f\left(1+\frac{y-1}{1-t(y-1)}\right)$
Thanks. I'll get back on this.

if we gather the terms above we get the same as Alan.
$u(y,t) = f \left( \xi(y,t) \right)$ where the characteristic  $\xi(y,t) = \frac{y + t - t y}{1 + t - t y}$.

All's well that end's well. Alan
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### Re: About solving a transport equation

$u=f\left(1+\frac{y-1}{1-t(y-1)}\right)$
Thanks. I'll get back on this.

I am working under the assumption that it is the solution of $\frac{\partial u}{\partial t} - \frac{\partial u}{\partial y} = 0$

No, Paul is just posting my answer again for the problem you posted.

It is kind of interesting to see what happens there for initial data on $y > 1$.
Suppose initial data $f(y)$ over the whole positive y-axis.
Then, picture $u(y,t)$ at fixed time-slice t for all $y>0$.

As we already have seen, for y=0 to y=1 in the time-slice plot, we just see a somewhat distorted image of $f(y)$ originally from $y=t/(t+1)$ to $y=1$, the fixed point. So, it is kind of like using a microscope to examine just part of the initial data (parts near $y=1$). As t gets larger, so does the power of the microscope.

For y>1, you see an image of  $f(y)$ originally from $y=1$ to $y=\infty$, squished into the time-slice interval $y=1$ to $y = y^*(t) \equiv 1 + 1/t$. So, this is the opposite idea: the whole range of the initial data is packed into a finite interval. Kind of like a trash compactor. As t gets larger, the size of the compactified trash gets smaller.

For $y > y^*(t)$, I think we are starting to examine initial data on $y < 0$, which wasn't defined by me (above), so probably best to say $u(y,t)$ is undefined.  